MA 162 - Quiz 10 11 12 - Solutions.pdf

# MA 162 - Quiz 10 11 12 - Solutions.pdf - MA 162 12...

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MA 162 - Quiz 10, 11, 12 Solutions Problem 1 (Quiz 10 - Problem 1.a) . Find all values of p for which the integral below converges ( Fall 2015 - Exam 2 - Problem 11 ) Z 1 dx x p +1 Solution. To solve this problem, we need to see how p effects the outcome of this integral. To do so we must evaluate the integral: Z 1 dx x p +1 = lim t →∞ Z t 1 dx x p +1 = lim t →∞ - 1 p 1 x p t 1 = lim t →∞ - 1 p 1 t p - 1 , ( * ) From this we can conclude that p 6 = 0. In the case where p > 0, 1 t p 0 as t → ∞ , then ( * ) becomes lim t →∞ - 1 p 1 t p - 1 = 1 p , so the integral is convergent. In the case p < 0, 1 t p = t - p , and when t → ∞ , t - p → ∞ . Thus we conclude that when p > 0, this integral is convergent and when p 0 the integral is divergent.

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MA 162 - Quiz 10, 11, 12 Solutions Problem 2 (Quiz 10 - Problem 1.b) . Find all values of p for which the integral below converges ( Spring 2015 - Exam 2 - Problem 7 ) Z 1 0 dx x p +1 Solution. Again, we must evaluate the integral: Z 1 0 dx x p +1 = lim t 0 + Z 1 t dx x p +1 = lim t 0 + - 1 p 1 x p 1 t = lim t 0 + - 1 p 1 - 1 t p , ( * ) From this we can conclude that p 6 = 0. In the case where p < 0, 1 t p = t - p 0 as t 0 + , then ( * ) becomes lim t 0 + - 1 p 1 - 1 t p = - 1 p , so the integral is convergent 1 . In the case p < 0, when t 0 + , 1 t p → ∞ . Thus we conclude that when p < 0, this integral is convergent and when p 0 the integral is divergent.
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